Calculate the power of the eye in D when viewing an object 4.18 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter your answer to at least one decimal place.)

A converging lens with a focal length of 40 cm and a diverging lens with a focal length of -40 cm are 150 cm apart. A 3.0-cm-tall object is 60 cm in front of the converging lens. Calculate the image height. Express your answer in centimeters to two significant figures. h = IVE ΑΣΦ ? cm

The projection lens in a certain slide projector is a single thin lens. A slide 23.4 mm high is to be projected so that its image fills a screen 1.84 m high. The slide-to-screen distance is 3.10 m. (Enter your answers to at least one decimal place.) (a) Determine the focal length of the projection lens. (b) How far from the slide should the lens of the projector be placed to form the image on the screen?

Two converging lenses, both with focal lengths 34.5 cm are placed a distance 78.3 cm apart. An object is placed 88.7 cm from the 1st lens. Where is the final image located relative to the 2nd lens? Let + be the back side and - be the front side regarding this distance from the 2nd lens. Measure this in cm.

An object of height 7.25 cm is placed at a distance of 22.5 cm from the convex lens. Whose focal length is 10.5 cm. Then Calculate the following: (a) The position of the image in cm from the lens: (b) The magnification of the lens (Only magnitude): (c) Height of the image formed:

1) Calculate the magnification of the 36-in refractor at Lick Observatory on Mount Hamilton near San Jose, California. The focal lenght of the objective lens is 17.37m, and the focal lenght of the eyepiece is 22 mm. (Express your answer to three significant figures.) 2) What is the significance of the negative sign in the magnification equation?       a) The image is inverted when viewed in the eyepiece.        b) The image is upright when viewed in the eyepiece.

Compute the focal length of the lens which will give an erect image 10cm from the lens when the object distance from the lens is (a) 200 cm (b) very great.

A converging lens has a focal length of 15.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.) (a)    15.0 cm q =  cm M =  Select all that apply to part (a).  realvirtualuprightinvertedno image (b)    5.00 cm q =  cm M =  Select all that apply to part (b).  realvirtualuprightinvertedno image (c)    45.0 cm q =  cm M =  Select all that apply to part (c).  realvirtualuprightinvertedno image

Assume that the near point of a hyperopic (farsighted) eye is 0.936 m in front of the eye. A lens is used to let the eye see clearly an object that is 0.250 m in front of the eye. Ignore the small distance from the lens to the eye. What is the power of the lens (in diopters; answer sign and magnitude)?

= 1.0). An object (A) is Assume that a thin lens (r, placed in the left side of lens (d = 45mm) and the image of the object is located in the right side of the lens at position (B) (d' = 90mm). 1) Compute the refraction index of the lens (n2); = -r2 = 40mm) is placed in the air (n 2) If placing another negative thin lens (f2 = -180mm) immediately behind the first lens to create a doublet lens in which the spacing between two lenses is zero (as shown in the right figure), what is the focal length of this doublet? 3) Where is the final image position (C) of the object (A) through this new doublet? (Note: Assuming this doublet lens is a combined thin lens too without considering its thickness).